Problem: Simplify the following expression: $\dfrac{8q^5}{32q^3}$ You can assume $q \neq 0$.
$ \dfrac{8q^5}{32q^3} = \dfrac{8}{32} \cdot \dfrac{q^5}{q^3} $ To simplify $\frac{8}{32}$ , find the greatest common factor (GCD) of $8$ and $32$ $8 = 2 \cdot 2 \cdot 2$ $32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ $ \mbox{GCD}(8, 32) = 2 \cdot 2 \cdot 2 = 8 $ $ \dfrac{8}{32} \cdot \dfrac{q^5}{q^3} = \dfrac{8 \cdot 1}{8 \cdot 4} \cdot \dfrac{q^5}{q^3} $ $\phantom{ \dfrac{8}{32} \cdot \dfrac{5}{3}} = \dfrac{1}{4} \cdot \dfrac{q^5}{q^3} $ $ \dfrac{q^5}{q^3} = \dfrac{q \cdot q \cdot q \cdot q \cdot q}{q \cdot q \cdot q} = q^2 $ $ \dfrac{1}{4} \cdot q^2 = \dfrac{q^2}{4} $